In this post I want to talk about an innocent commutative algebra lemma. Let be a DVR with uniformizer , and let be a finite torsion -module, so for some uniquely determined sequence . I’ll somewhat abusively refer to the ‘s as the “elementary divisors” of .

Lemma. If is an -submodule generated by elements, then . Furthermore, if equality holds, then is a direct summand of .

(Here and throughout, denotes -module length.)

Proof. We first prove the inequality by induction on . Fix a surjection such that . Choose a minimal basis of such that generates . Choose some with for all , and make the substitution for . Having done this, we get a basis with generating and for all . Let denote the -submodule generated by ; applying our induction hypothesis to the modules , we get the inequality . Since and generate , we get an inequality . But , since annihilates , so
as desired.

For the second claim, we argue by induction on ; the case is easy (argument: must project isomorphically onto a direct summand of of the form ). Maintain the previous notation, and assume we have an equality . Since and , the chain of inequalities
then forces and . Since and are generated by and element, respectively, they are both direct summands of by the induction hypothesis. Finally, the equality implies that , so is a direct summand of .

This lemma really really really looks like it should be well-known, but I couldn’t find it stated in the literature. Presumably I was just typing the wrong things into google. Can some reader provide a reference? If you can find a reference in a textbook (not a research paper), this will settle a bet between me and AJdJ. Also, it would be really nice if there were a “coordinate-free” proof which didn’t involve choosing a basis for . Before finding the argument given above, I spent a while trying to make a proof based on the theory of Fitting ideals; the latter seem quite natural in this context, since one has the equality for any finite -module. Can the reader make such a proof work?

OK literally while writing the previous two sentences I hit upon the following argument for the first part of the lemma. It clearly suffices to show the complementary inequality For this we use Fitting ideals as follows. Recall that for any finite torsion module over with elementary divisors , we have an equality ; in particular, , and if is generated by elements. Returning to the situation at hand, we have an inclusion (this is a special case of Proposition XIII.10.7 in Lang’s Algebra). But since by assumption is generated by elements, so we get
and this immediately implies the desired inequality.